Question

IMPOSSIBLE QUESTION:


Past records show that the times, in seconds, take to run 100m by children at a school can be modelled by a normal distribution witha mean of 16.12 and a standard deviation of 1.60.

On sports day the school awards certificates to the fastest 30% of the children in the 100m race.

Estimate, to 2d.p., the slowest time taken to run 100m for which a child will be awarded a certificate.

Answer 1

X~N(16.12, 1.60^2)

Answer 2

Quite possible.

Answer 3

what a smooth answer

Answer 4

Calculate the z score of the time that you're interested in.
Z-score = number of standard deviations from the mean.
(x-mean)/SD
Look that z score up on a z table, and that will tell you what percentage of students run slower or faster than that, depending on the table you're looking at.

Answer 5

Thank you, Muse.

Answer 6

Or, go backwards. Figure out what percent you need, find a z-score for that percent, then turn that into a time.

Answer 7

http://health.adelaide.edu.au/publichealth/staff/ASCIEB_statistical_tables.pdf
-Since we are concerned with the top 30% we want to use 'Table One' (One tailed probability) to find the zscore which fits 30% of the distribution to one end of the curve.
-In this case it is between .52 and .53, approximate or interpolate this value (~.5244)
-sub this z value into the equation that relates mean, SD, z-value, and data.
\[z = (x-u)/\sigma \]
where z is your z-value, u is your mean, \[\sigma\] is your standard deviation and x is your data point(which you need to solve for.

plug and play